c++ 編程練習 014:MyString

北大程序設計與算法（三）測驗題彙總(2020春季)

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描述

補足MyString類，使程序輸出指定結果

#include <iostream>
#include <string>
#include <cstring>
using namespace std;
class MyString {
	char * p;
public:
	MyString(const char * s) {
		if( s) {
			p = new char[strlen(s) + 1];
			strcpy(p,s);
		}
		else
			p = NULL;

	}
	~MyString() { if(p) delete [] p; }
// 在此處補充你的代碼
};
int main()
{
	char w1[200],w2[100];
	while( cin >> w1 >> w2) {
		MyString s1(w1),s2 = s1;
		MyString s3(NULL);
		s3.Copy(w1);
		cout << s1 << "," << s2 << "," << s3 << endl;

		s2 = w2;
		s3 = s2;
		s1 = s3;
		cout << s1 << "," << s2 << "," << s3 << endl;
		
	}
}

輸入
多組數據，每組一行，是兩個不帶空格的字符串

輸出
對每組數據，先輸出一行，打印輸入中的第一個字符串三次,然後再輸出一行，打印輸入中的第二個字符串三次

樣例輸入
abc def
123 456

樣例輸出
abc,abc,abc
def,def,def
123,123,123
456,456,456

來源
Guo Wei

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分析

MyString s1(w1)需要構造函數，題目已經給出；
MyString s2 = s1需要複製構造函數；
s2 = w2;需要臨時構造函數；
s3 = s2;s1 = s3;需要等號運算符的重載；

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解答

void Copy(const char * s){
		if(p)
			delete []p;
		if(s){
			p = new char[strlen(s) + 1];
			strcpy(p,s);
		}
		else
			p = NULL;
	}
	friend ostream &operator<<(ostream &o,const MyString & s){
		o << s.p;
		return o;
	}
	MyString (const MyString & s) {
		if( s.p ) {
			p = new char[strlen(s.p)+1];
			strcpy(p,s.p);
		}
		else {
			p = NULL;
		}
	}
	MyString &operator = (const MyString & s){
		if(this == &s)
			return *this;
		if(s.p)
		{
			if(p)
				delete []p;
			p = new char[strlen(s.p) + 1];
			strcpy(p,s.p);
		}
		else
			p = NULL;
		return *this;
	}

執行效果